Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

What's a simple proof for $^{t}(A \cdot B) = ^{t}B \cdot ^{t}A$ with $A \in M_{m,n}, B \in M_{n, p}$ ?

share|improve this question
1  
Is that transposition? –  J. M. Dec 30 '11 at 15:28
    
And is the $\cdot$ matrix multiplication? –  Henning Makholm Dec 30 '11 at 15:29
    
Yes to both your questions –  Mats Dec 30 '11 at 15:31
1  
$(AB)^T = B^T A^T$ -- matrices multiply by juxtaposition. –  Henning Makholm Dec 30 '11 at 15:47
2  
What about this? proofwiki.org/wiki/Transpose_of_Matrix_Product –  Martin Sleziak Dec 30 '11 at 16:24

4 Answers 4

up vote 7 down vote accepted

The $(i,j)$ entry of $(AB)^T$ is the $(j,i)$ entry of $AB$, $\sum \limits_k a_{jk}b_{ki}$. The $(i,j)$ entry of $B^TA^T$ is $\sum \limits_k b_{ki}a_{jk}$ (the $i$th row of $B^T$ is the $i$th column of $B$, the $j$th column of $A^T$ is the $j$th row of $A$).

Just a tip, you should remember and understand that the $(i,j)$ entry of multiplying $AB$ is $\sum \limits_k a_{ik}b_{kj}$. As $k$ advances, you are going across row $i$ of $A$ and down column $j$ of $B$.

share|improve this answer

For what's it's worth:

$$ \overbrace{\left[\matrix{a_{11}&a_{12}&a_{13} \cr \color{maroon}{a_{21}}&\color{maroon}{a_{22}}&\color{maroon}{a_{23}} \cr a_{31}&a_{32}&a_{33} \cr }\right]}^A \overbrace{ \left[\matrix{b_{11}&b_{12}&\color{maroon}{b_{13}} \cr b_{21}&b_{22}&\color{maroon}{b_{23}} \cr b_{31}&b_{32}&\color{maroon}{b_{33}} \cr }\right]}^B = \underbrace{ \left[\matrix{ \cdot & \cdot & \cdot \cr \cdot & \cdot & \color{maroon}{ a_{21}b_{13}+ a_{22}b_{23}+ a_{23}b_{33} } \cr \cdot & \cdot & \cdot \cr }\right]}_{AB} $$ $$ \phantom{ \overbrace{\left[\matrix{a_{11}&a_{12}&a_{13} \cr \color{maroon}{a_{21}}&\color{maroon}{a_{22}}&\color{maroon}{a_{23}} \cr a_{31}&a_{32}&a_{33} \cr }\right]}^A \overbrace{ \left[\matrix{b_{11}&b_{12}&\color{maroon}{b_{13}} \cr b_{21}&b_{22}&\color{maroon}{b_{23}} \cr b_{31}&b_{32}&\color{maroon}{b_{33}} \cr }\right]}^B=} \underbrace{ \left[\matrix{ \cdot & \cdot & \cdot \cr \cdot & \cdot & \cdot \cr \cdot & \color{maroon}{ a_{21}b_{13}+ a_{22}b_{23}+ a_{23}b_{33} } & \cdot \cr }\right]}_{[AB]^T} $$ $$ \underbrace{ \left[\matrix{b_{11}&b_{21}&b_{31} \cr {b_{12}}&{b_{22}}&{b_{32}} \cr \color{maroon}{b_{13}}&\color{maroon}{b_{23}}&\color{maroon}{b_{33}} \cr }\right]}_{B^T} \underbrace{ \left[\matrix{a_{11}&\color{maroon}{a_{21}}& {a_{31}} \cr a_{12}&\color{maroon}{a_{22}}& {a_{32}} \cr a_{13}&\color{maroon}{a_{23}}& {a_{33}} \cr }\right]}_{A^T} = \underbrace{ \left[\matrix{ \cdot & \cdot & \cdot \cr \cdot & \cdot & \cdot \cr \cdot & \color{maroon}{ a_{21}b_{13}+ a_{22}b_{23}+ a_{23}b_{33} } & \cdot \cr }\right]}_{B^TA^T} $$

share|improve this answer

Unfold the definitions of matrix multiplication and transposition in both cases. Notice that each element of the matrix on the two sides of the equals signs end up being the same sum of products of elements of $A$ and $B$.

share|improve this answer
1  
Could you give a worked out example, I mess up with the indexes. –  Mats Dec 30 '11 at 15:38

Let $r_1,.., r_n$ be the rows of $A$ and $c_1,..,c_n$ be the columns of $B$.

The $ij$ entry in $(AB)^T$ is the $ji$ entry in $AB$, which is the dot product $r_j \cdot c_i$.

The $ij$ entry in $B^TA^T$ is the dot product between the $i$th row of $B^T$ and $j$th column of $A^T$, thus the dot product between the $i$th column of $B$ and $j$th row of $A$, hence $c_i \cdot r_j$.

Since

$$r_j\cdot c_i =c_i \cdot r_j $$

we get

$$(AB)^T=B^TA^T \,.$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.